The components of the earth rotation at the latitude $\lambda$ are:
$$\mathbf \Omega_E=\begin{bmatrix} 0\\ \cos(\lambda)\, \Omega\\ \sin(\lambda)\,\Omega\\ \end{bmatrix}$$
from here we can obtain the pseudo forces due to the earth rotation
$$ \mathbf F_s=m\,\big(2\,(\mathbf\Omega_E\,\times \mathbf{\dot{R}})+\mathbf\Omega_E\times\,(\mathbf\Omega_E\times \mathbf R)\big) $$
where $$\mathbf R=\begin{bmatrix} x \\ y \\ z\\ \end{bmatrix}$$ is the postion vector$$\mathbf R=\begin{bmatrix} x \\ y \\ z\\ \end{bmatrix}\quad,\text{is the postion vector}$$
$$\frac{\mathbf F_s}{m}= \left[ \begin {array}{c} \left( \left( \cos \left( \lambda \right) \right) ^{2}x+ \left( \sin \left( \lambda \right) \right) ^{2}x \right) {\Omega}^{2}+ \left( -2\, \cos \left( \lambda \right) { \dot{z}}+2\, \sin \left( \lambda \right) {\dot{y}} \right) \Omega \\ \left( - \sin \left( \lambda \right) \cos \left( \lambda \right) z+ \left( \sin \left( \lambda \right) \right) ^{2}y \right) {\Omega}^{2}-2\, \sin \left( \lambda \right) \Omega\,{\dot{x}}\\ \left( \left( \cos \left( \lambda \right) \right) ^{2}z- \cos \left( \lambda \right) \sin \left( \lambda \right) y \right) {\Omega}^{2}+2\, \cos \left( \lambda \right) \Omega\,{\dot{x}}\end {array} \right] $$
The EOM's in rotating system free falling
$$\begin{bmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{z} \\ \end{bmatrix}=-\frac{\mathbf F_s}{m}-\begin{bmatrix} 0 \\ 0 \\ g \\ \end{bmatrix}\tag 1$$$$\begin{bmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{z} \\ \end{bmatrix}=-\frac{\mathbf F_s}{m}-\begin{bmatrix} 0 \\ 0 \\ g \\ \end{bmatrix}\tag 1\\ g=\frac{G\,M_E}{(R_E+z)^2}=\frac{M_E}{\big(R_E\,(1+\frac{z}{R_E})\big)^2 }\approx \frac{G\,M_E}{R_E^2}$$ where $~R_E~$ is the earth radius , $~M_E~$ earth mass and G the gravitation constant
with $~\Omega=7.2710^{-5}~$[1/s]$~~,\lambda=\frac{40}{180}\,\pi~$
$$\frac{\mathbf F_s}{m}=\left[ \begin {array}{c} 0.000000005285290000\,x- 0.0001113828620\,{ \dot z}+ 0.00009346131845\,{\dot y}\\ - 0.000000002602497284\,z+ 0.000000002183754512\,y- 0.00009346131845\,{ \dot x}\\ 0.000000003101535488\,z- 0.000000002602497284\,y+ 0.0001113828620\,{\dot x}\end {array} \right] $$
thus $\frac{\mathbf F_s}{m}\approx \mathbf 0~$ and equation (1) is the inertial case.